Problem 21.63 A proton follows the path shown in (Figure 1). Its initial speed is v0 = 1.4 x 10^6 m/s. Part A What is the proton’s speed as it passes through point P? Express your answer to two significant figures and include the appropriate units.
Answer for "What is the proton’s speed as it passes through point P?"
Initial speed of the proton:
v_{0}=1.4 \times 10^{6} \mathrm{~m} / \mathrm{s}
\begin{aligned} \frac{1}{2} m v_{0}^{2}+\frac{k Q q}{r_{1}} &=\frac{k Q q}{r_{2}}+\frac{1}{2} m v^{2} \\ \frac{1}{2} m v^{2} &=\frac{1}{2} m v_{0}^{2}+\frac{k Q q}{r_{1}}-\frac{k Q q}{r_{2}} \\ &=\left[\frac{1}{2} \times 1.67 \times 10^{-27} \times\left(1.4 \times 10^{6}\right)^{2}\right]+\left[\frac{9 \times 10^{9} \times 1.6 \times 10^{-19} \times\left(-10 \times 10^{-9}\right)}{3 \times 10^{-3}}\right]-\left[\frac{9 \times 10^{9} \times 1.6 \times 10^{-19} \times\left(-10 \times 10^{-9}\right)}{4 \times 10^{-3}}\right] \\ &=\left(1.64 \times 10^{-15}\right)-\left(4.8 \times 10^{-15}\right)+\left(3.6 \times 10^{-15}\right) \\ \frac{1}{2} m v^{2} &=0.44 \times 10^{-15} \\ v &=\sqrt{\frac{2 \times 0.44 \times 10^{-15}}{1.67 \times 10^{-2)}}}=7.3 \times 10^{3} \mathrm{~m} / \mathrm{s} \end{aligned}Hence the final speed of the proton is
7.3 \times 10^{5} \mathrm{~m} / \mathrm{s}